Electrical Circuits 2.1 INTRODUCTION CHAPTER

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1 CHAPTE Electrcal Crcuts. INTODUCTION In ths chapter, we brefly revew the three types of basc passve electrcal elements: resstor, nductor and capactor. esstance Elements: Ohm s Law: The voltage drop across a lnear resstor s proportonal to the current flowng through the resstor, where the constant of proportonalty s the resstance as shown n Fg... Fg.. = esstors do not store electrcal energy n any form but dsspate t as heat. The rate of energy dsspated (power consumed) by a resstor s gven by P = = (W or J/s) esstors n Seres: The current passes through each element as shown n Fg..(a). The total voltage drop s gven by the sum of the voltage drops across each element, or = Applyng Ohm s law, we obtan eq = or eq = The voltage drop across each resstor s then gven by =

2 98 MATLAB: An Introducton wth Applcatons and = (a) esstors n seres Fg.. (b) esstors n parallel esstors n Parallel: All the elements n the case have the same voltage appled across them as shown n Fg.. (b). = Applyng Ohm s law, we get or = eq eq = If there are n resstors, we can wrte... = eq n Solvng eq. (.) for the currents and, we obtan = =...(.) Inductance Elements: The voltage L across the nductor L s gven by (see Fg..3) L = d L L dt L L L Fg..3 where L s the current through the nductor.

3 Electrcal Crcuts 99 The energy stored n an nductance s E = L = L ( q ) where q s the electrcal charge. The voltage drop n an nductance s = L = L q Inductances n Seres: Snce the voltage drop through an nductor s proportonal to the nductance L, we have (Fg..4 (a)) L eq = L L L L L (a) Inductances n seres Fg..4 Inductances L (b) Inductances n parallel Inductances n Parallel: eferrng to Fg..4 (b), we have L eq = Smlarly, for n nductors LL L L... = Leq L L L n Capactance Elements: Capactance C s a measure of the quantty of charge that can be stored for a gven voltage across the plates. The capactance C of a capactor s gven by q C = c where q s the quantty of charge stored and c s the voltage across the capactor. Snce = dq dt and c = q/c, we have d = C c dt or d c = C dt Hence c = C c dt

4 00 MATLAB: An Introducton wth Applcatons Ths s shown n Fg..5. C Fg..5 Capactor The energy stored n a capactor s gven by E = C The voltage drop across capactor s gven by c = q = C Capactors n Seres: dt C eferrng to Fg..6 (a) we have C For n capactors eq CC = C C... = Ceq C C C n Capactors n Parallel: For capactors n parallel (see Fg..6 (b)) C eq = C C c C C C (a) Capactors n seres (b) Capactors n parallel Fg..6 Capactor. ELECTICAL CICUITS In ths secton, we apply Ohm s law to seres and parallel crcuts to determne the combned resstance of the gven crcut. Seres Crcuts: The combned resstance of seres-connected resstors of a smple seres crcut s gven by the sum of the ndvdual resstances. The voltage between ponts A and B of the smple seres crcut shown n Fg..7 s gven by

5 Electrcal Crcuts 0 A 3 B Fg..7 Seres crcut 3 where = = and 3 = 3 = 3 Hence = 3 Therefore, the combned resstance of the seres crcut s gven by = 3 Parallel Crcuts: For the parallel electrcal crcut shown n Fg..8, we can wrte = = and 3 = Now = 3 Therefore, = = 3 where s the combned resstance. Hence = 3 Fg..8 Parallel crcut

6 0 MATLAB: An Introducton wth Applcatons Therefore, 3 = = KICHHOFF S LAWS Krchhoff s laws are the two most useful physcal laws for modelng electrcal systems. It s necessary to apply Krchhoff s laws n solvng electrc crcut problems as they nvolve many electromotve forces such as resstance, capactance and nductance. The Krchhoff s laws are stated as follows:. Krchhoff s current law (node law): The algebrac sum of all the currents flowng nto a juncton (or node) s zero (node analyss). In other words, the sum of currents enterng a node s equal to the sum of the currents leavng the same node. A node s an electrcal crcut s a pont where three or more wres are joned together. Currents gong toward a node are consdered postve whle currents leavng a node are treated as negatve. The algebrac sum of all currents (n) a crcut node s zero. That s, ( ) = 0 n n n eferrng to Fg..9, Krchhoff s current law states that (a) (b) (c) (d) Fg..9 Fg..9 (a) 3 =0 Fg..9 (b) ( 3 )=0 Fg..9 (c) 3 =0 Fg..9 (d) =0. Krchhoff s voltage law (loop law): The algebrac sum of all the potental drops around a closed loop (or closed crcut) s zero (loop analyss). In other words, the sum of the voltage drops s equal to the sum of the voltage rses around a loop. That s, the sum of all voltage drops around a crcut loop s zero. Hence Σ drop = 0 or Σ gan = 0

7 Electrcal Crcuts 03 The voltage drops or voltage gans should be approprately ndcated for loop analyss. Fgure.0 shows examples wth useful sgn conventon. A A E B e AB = E (a) (b) B e AB = A A L B E B a c e AB = E e AB = (c) (d) (e) Fg..0 The applcaton of MATLAB to the analyss and desgn of control systems, engneerng mechancs (statcs and dynamcs), mechancal vbraton analyss, electrcal crcuts and numercal methods s presented n ths chapter wth a number of llustratve examples. The MATLAB computatonal approach to the transent response analyss, steps response, mpulse response, ramp response and response to the smple nputs are presented. Plottng root loc, Bode dagrams, polar plots, Nyqust plot, Nchols plot and state space method are obtaned usng MATLAB. Extensve worked examples are ncluded wth a sgnfcant number of exercse problems to gude the student to understand and as an ad for learnng about the analyss and desgn of control systems, engneerng mechancs, vbraton analyss of mechancal systems, electrcal crcuts, and numercal methods usng MATLAB..4 EXAMPLE POBLEMS AND SOLUTIONS Example E.: Fgure E. shows an electrcal crcut wth resstors and voltage sources. Wrte a MATLAB program to determne the current n each resstor usng the mesh current method based on Krchhoff s voltage law. Gven: =, =, 3 = 44, = 0Ω, = Ω, 3 = 5 Ω, 4 = 7 Ω, 5 = 6 Ω, 6 = 0 Ω, 7 = 0 Ω, 8 = 5 Ω

8 04 MATLAB: An Introducton wth Applcatons Fg. E. Soluton: Let,, 3 and 4 be the loop currents as shown n Fg. E.. Accordng to Krchhoff s voltage law: sum of voltage around closed crcut s zero. Thus, the loop equatons can be wrtten by takng n each loop clockwse drecton as reference. 3 ( 3 ) ( ) = 0 ( ) 4 ( 3 ) 7 ( 4 ) 5 = 0 3 ( 3 ) 6 ( 3 4 ) 4 ( 3 ) = ( 4 ) 6 ( 4 3 ) = 0 The equatons can be wrtten n matrx form as follows: ( ) ( ) ( ) = ( ) MATLAB soluton of ths system of equatons s gven below: MATLAB Program %INITIALIZING THE AIABLES =; =; 3=44; =[;0;;3]%CEATE THE OLTAGE ECTO =0; =; 3=5; 4=7; 8

9 Electrcal Crcuts 05 5=6; 6=0; 7=0; 8=5; % CEATE THE ESISTANCE MATIX =[(3) 3 0; (457) 4 7; 3 4 (346)6; (678)]; % GET THE CUENT ECTO AS SOLUTION I=nv()*; % ALLOT ALUES TO FOU CUENTS =I() =I() 3=I(3) 4=I(4) The output obtaned s as follows: = 0 44 = = = =.6655 The current n resstor = = 0.63 A The current n resstor 3 = 3 = A The current n resstor 4 = 3 = A The current n resstor 6 = 4 3 = A The current n resstor 7 = 4 = A

10 06 MATLAB: An Introducton wth Applcatons Example E.: Wrte a MATLAB program that computes the voltage across each resstor and the power dsspated n each resstor for the crcut shown n Fgure E. that has resstors connected n seres. 3 v S Fg. E. The voltage across each of the several resstors connected n seres s gven by the voltage dvder rule n vs v s v s v s v n v n = eq where v n, n = the voltage across resstor n and ts resstance, eq = n = the equvalent resstance, v s = the source voltage. The power dsspated n each resstor s gven by P n = n vs eq Soluton: MATLAB program s gven for the followng data: s =, = 0 Ω, = 7 Ω, 3 = 6 Ω, 4 = 9 Ω, 5 = 4 Ω, 6 = 7.5 Ω, 7 = 0 Ω % THIS POGAM CALCULATES THE OLTAGE ACOSS EACH ESISTO % IN A CICUIT THAT HAS ESISTOS CONNECTED IN SEIES vs=nput( Enter the source voltage ); rn=nput( Enter values of resstors as elements n a row vector\n ); req=sum(rn); % CALCULATING EQUIALENT ESISTANCE vn=rn*vs/req; % APPLY OLTAGE DIIDE ULE pn=rn*vs^/req/; % CALCULATING POWE IN EACH CICUIT =vs/req; % CALCULATE CUENT IN THE CICUIT ptotal=vs*; % CALCULATE POWE IN THE CICUIT table=[rn,vn,pn ];% CEATE TABLE dsp( esstance oltage Power ) %DISPLAY HEADINGS FO COLUMNS dsp( (ohms) (volts) (watts) ) dsp(table) % DISPLAY THE AIABLE TABLE